ccassel wrote:x^n = x^(n+2) for any integer n. Is it true that x>0?
(1) x = x^2 -2
(2) 2x < x^5
Only three values of x will satisfy the condition that x^n = x^(n+2) for any integer n:
x=-1, x=0, x=1.
Let's plug x=-1, x=0, and x=1 into x^n = x^(n+2).
If n=2 and n+2=4:
(-1)² = (-1)�. Yes.
0² = 0�. Yes.
1² = 1�. Yes.
If n=3 and n+2=5:
(-1)³ = (-1)�. Yes.
0³ = 0�. Yes.
1³ = 1�. Yes.
Thus, our only options are x=-1, x=0, or x=1.
Since we want to know whether x>0, the question can be rephrased: Does x=1?
Statement 1: x = x² - 2.
x² - x - 2 = 0.
(x-2)(x+1) = 0.
x=2 or x=-1.
Thus, it is not true that x=1.
Sufficient.
Statement 2: 2x < x�.
Of the 3 possible values x=-1, x=0, and x=1, only x=-1 satisfies statement 2:
2(-1) < (-1)�
-2 < -1.
Thus, it is not true that x=1.
Sufficient.
The correct answer is
D.
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